I did the test and the answer is C.
44.0095 you're welcome hope this helps
Answer:
1.5 moles
Explanation:
The equation of the reaction is given as:
2 MnO2 + 4 KOH + O2 --> 2KMnO 4 + 2KOH + H2
From the equation,
2 moles of MnO2 produces 2 moles of KMnO4
x moles of MnO2 would produce 1.5 moles of KMnO4
2 = 2
x = 1.5
Solving for x;
x = 1.5 * 2 / 2
x = 1.5 moles
Answer:
a) T
b) T
c) F
d) F
e) T
f) T
g) T
h) F
I) F
j) F
k) F
l) F
Explanation:
The w/v concentration is obtained from, mass/volume. Hence;
%w/v= 50/1000= 5%
In the %w/w we have;
25g/100 g = 25% w/w
In combustion reaction, energy is given out hence it is exothermic.
Neutralization reaction yields a salt and water
% by mass of carbon is obtained from;
8× 12/114 × 100 = 84.1%
All the ionic substances mentioned have very low solubility in water.
One mole of a substance contains the Avogadro's number of each atom in the compound.
There are two iron atoms so one mole contains 2× 55.85 g of iron.
Some sulphates such as BaSO4 are insoluble in water.
Halides are soluble in water hence NaI is soluble in water.
The equation does not balance with the given coefficients because the number of atoms of each element on both sides differ.
The equation represents a decomposition of calcium carbonate as written.
Iodine is decolorized.
The first reaction stated in the question occurs as follows;
2 KI (aq) + 2 H2SO4 (aq) + MnO2 (s) → MnSO4 (aq) + K2SO4 (aq) + I2 (s) + 2 H2O (l)
The reaction here is the formation of iodine from MnO2 and KI in the presence of dropwise H2SO4.
Hypo is the common name of sodium thio-sulphate or sodium hypo-sulfite.
The equation of the titration reaction is;
2Na2S2O3 + I2→ Na2S4O6 + 2NaI
When this reaction takes place, iodine is decolorized due to its reduction to I^-.
